
It can be easily shown that \(\frac{\rho ^2}{2\sigma ^2}=\frac{\pi e \Delta \gamma }{4}\) .
For \(\Delta \rightarrow \infty \) , the term \(\tau q^{4\rho ^2}\)  dominates the sum in \(\Theta _{\Lambda }(q)\)  [1]}. As proven in Appendix REF , (REF ) becomes
\(P_e(ub) =~\frac{\tau }{2} \exp (-\frac{\pi e \Delta \gamma }{4})+ o\left(\exp (-\frac{\pi e \Delta \gamma }{4})\right).\) 
